A long rectangular sheet of metal, 12cm wide, is to be made into a rain gutter by turning up two sides which make an angle of 120 degrees with the base. How many cm should be turned up to give the gutter its greatest capacity?

Question

Emily Ammerman · Accepted Answer

First, draw a sketch of the problem
This is an isosceles trapezoid with equal sides BC and AD #=x# The length base CD #=b_1=12-2x# If you draw a perpendicular from C to BA, that will be the height #h# #h=xsin60# Length of base BA #=b_2=(12-2x)+2xcos60# Area of trapezoid #=(1/2)(b_1+b_2)(h)# Insert the known values into the area formula ... Area#=(1/2)(12-2x+12-2x+2xcos60)(xsin60)# Area#=(12x-3/2x^2)(sqrt3/2)# Now, take the derivative with respect to x and set equal to zero ... #y'=(12-3x)(sqrt3/2)=0# Solution : #x=4# cm

William Abdalla · Answer

undefined To find the width to be turned up to give the gutter its greatest capacity, follow these steps:

1. Let $ x $ represent the width to be turned up on each side.
2. The length of the base of the gutter after turning up the sides will be $ 12 - 2x $.
3. The height of the gutter will be $ x $ (since the sides are turned up at an angle of 120 degrees).
4. The cross-sectional area of the gutter can be found by multiplying the length and the height.
5. To find the maximum capacity, differentiate the expression for the cross-sectional area with respect to $ x $, set the derivative equal to zero, and solve for $ x $.
6. Once you find the value of $ x $, substitute it back into the expression for the cross-sectional area to find the maximum capacity.